Suppose (X, Y ) is a random vector deﬁned on some probability space (Ω, F, IP) with IE(X 2 ) < ∞, IE(Y 2 ) < ∞. The following implications are true. IE(X|Y ) = IE(X) ր X and Y are independent ց IE(Y |X) = IE(Y ) For a proof of the ﬁrst statements on the left above, assume X and Y are independent. Then the constant IE(X) is measurable with respect to σ(Y ). Since X and σ(Y ) are independent, for any G ∈ σ(Y ) we have IE(X)dIP = IE(X)IE(IG ) = IE(XIG ) by independence

G

ց X and Y are uncorrelated ր (1A)

=

G

XdIP

Thus IE(X|Y ) = IE(X). Similarly we have IE(Y |X) = IE(Y ). Next we prove the statements on the right side of (1A). Suppose that IE(Y |X) = IE(Y ). Using the tower property and then taking out what is known, we have IE(XY ) = IE(IE(XY |X)) = IE(X(IE(Y |X)) = IE(X)IE(Y ). Thus X and Y are uncorrelated. Similarly, IE(X|Y ) = IE(X) implies that X and Y are uncorrelated. However, the converse implications are not true in general. Counterexample 1: X and Y uncorrelated does not imply IE(X|Y ) = IE(X) Let Ω = {−1, 0, 1} with P({ω}) = 1/3 for each ω ∈ Ω. Let Y (ω) = ω and X(ω) = I{0} (ω). Then XY = 0 so IE(XY ) = 0. Also, IE(Y ) = 0, and so IE(XY ) = 0 = IE(X)IE(Y ); that is, X and Y are uncorrelated. However, since X is measurable with respect to σ(Y ) we have IE(X|Y ) = X which is never equal to IE(X) = 1/3. Counterexample 2: IE(X|Y ) = IE(X) does not imply that X and Y are independent Consider the independent random variables X and S of Exercise 3.2.12. Then IE(SX|X) = XIE(S|X) = XIE(S) = 0. However, as shown in Exercise 3.2.12, X and SX are not independent.